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The vacuum Einstein field equations with positive cosmological constant admit a closed spherical slicing with scale factor $a(t)=\ell \cosh [(t-t_0)/\ell ]$. The time-symmetric slice is at $t=t_0$. The substitution $T=2\arctan (e^{\tau })-\pi /2$ gives $dT/d\tau =\mathrm{sech}\tau$ and $\cos T=\mathrm{sech}\tau$, so the metric is ${\ell }^2{\sec }^{2}T[-d{T}^2+d{\Omega }_3^2]$. The conformal metric is a finite-time slab of the Einstein static universe, with spacelike conformal infinities $T=\pm \pi /2$.

For positive spatial curvature and positive cosmological constant, the effective potential has its maximum atThe critical case $V(a_E)=-1$ hasand contains the Einstein static universe.

With nonnegative conserved pressureless matter and radiation in cosmology, set $M=(8\pi G/3){\rho }_m{a}^3$, $Q=(8\pi G/3){\rho }_r{a}^4$, and $L=(8\pi G/3){\rho }_{\Lambda }$. The Friedmann equation becomes $(da/dt{)}^2+V(a)=-k$. For negative vacuum density the potential rises strictly from negative infinity to positive infinity, so an expanding branch has one maximum scale factor. For positive vacuum density it has one maximum: energy above that maximum gives monotonic expansion or contraction, energy below it gives disjoint recollapsing and bouncing branches, and equality admits an Einstein static universe together with asymptotic monotonic branches.

In ordinary spherical coordinates, $t\in \mathbb{R}$, $r\in [0,\infty )$, $\theta \in [0,\pi ]$ and $\varphi \in [0,2\pi )$ with periodic identification. The regular coordinate chart excludes the origin and polar axes; the endpoint ranges include the usual spherical identifications. With null coordinates $u=t-r$ and $v=t+r$, both are real and their joint domain is $v\ge u$. The inverse relations giveThe arctangent coordinates obey $-\pi /2<p\le q<\pi /2$. Since $du={\sec }^{2}pdp$ and $dv={\sec }^{2}qdq$, the rescaled metric isNow $T=p+q$, $R=q-p$, so $4dpdq=d{T}^2-d{R}^2$. Thus the Minkowski conformal compactification hasThe full coordinate ranges are linked, not independent: $T\in (-\pi ,\pi )$ and $R\in [0,\pi )$ with $|T|+R<\pi$. Inverse coordinates are $p=(T-R)/2$, $q=(T+R)/2$, $t=[\tan q+\tan p]/2$, $r=[\tan q-\tan p]/2$. The conformal factor isThe unphysical metric extends as the unit-radius Einstein static universe on $\mathbb{R}\times S^3$, where $0\le R\le \pi$ with the endpoint two-spheres collapsed. Physical Minkowski spacetime occupies the triangular radial domain $R\ge 0$, $|T|+R<\pi$; the edge $R=0$ is the ordinary spatial center, not an infinity or physical boundary.

In the closure, the infinite-coordinate limits have locationsThese are future timelike infinity, past timelike infinity, spacelike infinity, future null infinity and past null infinity. For example ${\mathcal{I}}^{+}$ has $v\to +\infty$ with finite $u$, while ${\mathcal{I}}^{-}$ has $u\to -\infty$ with finite $v$. A diagram point with $0<R<\pi$ stands for a two-sphere, and each null boundary is topologically $\mathbb{R}\times S^2$; their three corners are points after the angular collapse. On the null boundaries $\Omega =0$ and $d\Omega \ne 0$, whereas the three corners have a degenerate zero gradient. Thus a compactifying metric can be smooth there even though those infinity points are not ordinary smooth null-boundary points.

The geodesic endpoints in Minkowski conformal compactification can be established directly. A physical inertial timelike geodesic has $t=t_0+E\tau$, $\mathbf{x}={\mathbf{x}}_0+\mathbf{p}\tau$, with $E>0$ and $E^2-|\mathbf{p}{|}^2=1$. At future infinity $u,v\to +\infty$, so $p,q\to \pi /2$ and the curve ends at $i^{+}$. At past infinity both tend to $-\infty$, giving $i^{-}$. Every complete inertial timelike path has these endpoints, regardless of velocity.

For a future-directed null geodesic, $t=t_0+E\lambda$, $\mathbf{x}={\mathbf{x}}_0+E\mathbf{n}\lambda$ with $|\mathbf{n}|=1$. As $\lambda \to +\infty$, $v\to +\infty$ but $u$ has a finite limit; the endpoint lies on ${\mathcal{I}}^{+}$. As $\lambda \to -\infty$, $u\to -\infty$ but $v$ has a finite limit, giving an endpoint on ${\mathcal{I}}^{-}$. A conformal transformation preserves these null paths as unparametrized geodesics. The null affine parameter under a conformal rescaling obeys $d\tilde{\lambda }/d\lambda ={\Omega }^2$; here ${\Omega }^2=O({\lambda }^{-2})$ at the null endpoints, so the unphysical affine parameter reaches them in finite time, although the physical affine parameter is infinite.

A physical spacelike geodesic has $t=t_0+a\lambda$, $\mathbf{x}={\mathbf{x}}_0+\mathbf{b}\lambda$, with $|\mathbf{b}|>|a|$. At either end $r>|t|$ asymptotically, $u\to -\infty$ and $v\to +\infty$, so both ends approach $i^0$. Timelike and spacelike physical geodesic paths are not generally geodesics of the rescaled metric. Their physical proper time or length is infinite, while $\Omega =O(|\lambda {|}^{-2})$ gives finite unphysical proper time or length near the endpoints. Infinity at finite diagram coordinates does not imply incompleteness of physical Minkowski geodesics.

The original Penrose diagram suppresses the angular spheres. The displayed physical paths illustrate their distinct endpoint types; the two sloping boundaries are null infinity, while the left edge is the regular center.

For an Einstein static universe, both $a_t$ and $a_{tt}$ vanish. The Friedmann acceleration equation givesand $a_E>0$ therefore impliesThe Friedmann equation additionally fixes the spatial curvature:Thus the acceleration balance is necessary but must be accompanied by the static energy constraint.

Set $L=8\pi G{\rho }_{\Lambda }/3>0$. The Friedmann effective potential for matter, radiation and vacuum now readsWith nonzero nonnegative dust or radiation, $V$ tends to $-\infty$ at both ends. Its stationary point satisfiesEquivalently $V^{\prime }(a)=0$ when $2L=M/a^3+2Q/a^4$; the right-hand side decreases strictly, so there is exactly one stationary point. It is a maximum becauseThe energy line in the preceding sketch therefore gives five broad types of history:

At the critical energy there are also marginal branches within the monotonic classes: expansion from zero can approach $a_E$ asymptotically, or expansion can start asymptotically at $a_E$ and proceed to infinity; contraction gives their reversals. They must not be mistaken for reaching the static solution in finite time. Indeed near the maximum,so $dt\simeq da/[\sqrt{-V^{\prime \prime }(a_E)/2}|a-a_E|]$ diverges logarithmically. The static equilibrium is unstable. Since $V(a_E)<0$, the recollapsing, bouncing and static cases require positive spatial curvature $k>0$.

The physical spherical coordinates cover $t\in \mathbb{R}$, $r\in [0,\infty )$, $\theta \in [0,\pi ]$, and $\varphi \in [0,2\pi )$ with periodic identification. The center and the polar axes are the usual coordinate singularities, rather than physical curvature singularities; a regular polar chart restricts $r>0$ and $0<\theta <\pi$. For the null coordinates,Thus the physical metric tensor isThe conformal transformation followed by $p=\arctan u$, $q=\arctan v$ gives $du=(1+u^2)dp$, $dv=(1+v^2)dq$, andConsequently $d{s}^2=4dpdq-{\sin }^2(q-p)d{\Sigma }^2$. Finally $T=p+q$, $R=q-p$ converts $4dpdq$ to $d{T}^2-d{R}^2$, yieldingThe ranges must retain their joint constraints:Equivalently, $-\pi <T-R<\pi$ and $-\pi <T+R<\pi$, with $R\ge 0$. The angles retain their original ranges and identifications. Extending $R$ to negative values would double-count the radial polar geometry. The conformal factor isThe unphysical metric tensor extends onto the Einstein static universe, while physical Minkowski spacetime occupies the triangular region in its radial Penrose diagram. The edge $R=0$ is the regular center and belongs to the physical region away from its endpoints.

The conformal boundary consists of the following infinities. Future null infinity has $q=\pi /2$, hence $T+R=\pi$ with $0<R<\pi$, and corresponds to finite $u$ with $v\to +\infty$. Past null infinity has $p=-\pi /2$, hence $T-R=-\pi$, and corresponds to finite $v$ with $u\to -\infty$. Future timelike infinity is $(T,R)=(\pi ,0)$, past timelike infinity is $(-\pi ,0)$, and spacelike infinity is $(0,\pi )$. Although angles label points on null infinity, their two-spheres collapse at these three endpoint locations in the compactified picture.

For the geodesic endpoints in Minkowski conformal compactification, one can check the limits directly. A timelike inertial trajectory is $\mathbf{x}=\mathbf{b}+\mathbf{w}t$ with $|\mathbf{w}|<1$. At $t\to +\infty$ both $t-r$ and $t+r$ tend to $+\infty$, so $p,q\to \pi /2$ and the trajectory ends at future timelike infinity. At $t\to -\infty$ both tend to $-\infty$, giving past timelike infinity. This includes a stationary inertial worldline with bounded $r$.

For a null inertial trajectory, $|\mathbf{w}|=1$. As $t\to +\infty$, $r=t+\mathbf{w}\cdot \mathbf{b}+o(1)$, so $u$ has a finite limit while $v\to +\infty$; its future endpoint lies on future null infinity. As $t\to -\infty$, $r=-t-\mathbf{w}\cdot \mathbf{b}+o(1)$, giving a finite limiting $v$ and $u\to -\infty$, hence a past endpoint on past null infinity. Radial null geodesic paths have $T\pm R$ constant. A ray passing through the center changes its angular direction; in the angle-suppressed radial diagram it reflects at $R=0$ rather than terminating there.

A spacelike geodesic is also a straight line, now parametrized so that $|d\mathbf{x}/d\lambda |>|dt/d\lambda |$. At either end, $r>|t|$ asymptotically, $u\to -\infty$, $v\to +\infty$, and the endpoint is spacelike infinity. A radial spacelike line at $t=0$ has two opposite-angle halves which project onto the same horizontal interval in the diagram.

The conformal compactification endpoints occur at finite $T$ but at infinite physical proper time or affine parameter; Minkowski spacetime is geodesically complete. The conformal preservation of null geodesic paths does not preserve an affine parameter: if $\lambda$ is physically affine, an unphysical affine parameter satisfies $d\tilde{\lambda }/d\lambda ={\Omega }^2$ up to a constant. Along a complete null ray this can have a finite integral at infinity. Timelike and spacelike physical geodesic paths are not generally geodesics of the rescaled metric tensor, so their drawn curves must not be interpreted as freely falling unphysical observers.

Finally, every physical event can send an outgoing null ray to future null infinity, so $J^{-}({\mathcal{I}}^{+})$ is all of Minkowski spacetime. Its black-hole region and intrinsic event horizon are empty. The null edges in the diagram are infinity, not horizons concealing part of the physical manifold. Complete inertial observers likewise have no observer event horizon. Accelerated observers restricted to a Rindler wedge can have a Rindler horizon, but that is observer-dependent and does not contradict the horizon-free Minkowski spacetime. The conformal diagram compactifies infinity; it does not introduce a physical horizon.

The original spherical coordinates have $t\in \mathbb{R}$, $0\le r<\infty$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$, with $\varphi$ periodic. The angular coordinates degenerate at the polar axis and all angles identify at $r=0$; a regular spherical chart uses $r>0$, $0<\theta <\pi$ and suitable overlapping angular patches.

The retarded time $u=t-r$ and advanced coordinate $v=t+r$ both range over $\mathbb{R}$, with the joint restriction $u\le v$. Their inverse relations are $t=(u+v)/2$ and $r=(v-u)/2$, soSet $p=\arctan u$, $q=\arctan v$. ThenThe conformal factor is $\Omega =2\cos p\cos q>0$. Since $(\tan q-\tan p)\cos p\cos q=\sin (q-p)$, multiplication by ${\Omega }^2$ givesFinally $T=p+q$ and $R=q-p$ give $4dpdq=d{T}^2-d{R}^2$. HenceThis is a region of the Einstein static universe, with positive time and negative spatial signature. The exact joint coordinate domain isor equivalently $R\ge 0$ and $|T|+R<\pi$. The regular radial chart has $R>0$; $R=0$ is the ordinary centre. The angular ranges are unchanged. It is important not to treat the stated individual ranges of $T,R$ as independent. The inverse map and conformal factor provide a useful check:

The resulting Minkowski conformal compactification has a triangular radial Penrose diagram. The two sloping boundary edges areThese are future null infinity and past null infinity. The three vertices are future timelike infinity $i^{+}=(\pi ,0)$, past timelike infinity $i^{-}=(-\pi ,0)$ and spacelike infinity $i^0=(0,\pi )$. They are ideal boundary points, not events of physical Minkowski spacetime.

For a complete inertial timelike geodesic, $\mathbf{x}=\mathbf{b}+\mathbf{v}t$ with $|\mathbf{v}|<1$. As $t\to +\infty$, both $t-r$ and $t+r$ tend to $+\infty$, so $p,q\to \pi /2$ and the curve ends at $i^{+}$. As $t\to -\infty$, both tend to $-\infty$, giving $i^{-}$. These endpoints occur at infinite physical proper time.

For a complete null geodesic, write $\mathbf{x}=\mathbf{b}+\mathbf{n}t$, $|\mathbf{n}|=1$. To the future, $r=t+\mathbf{n}\cdot \mathbf{b}+O(1/t)$, so $u\to -\mathbf{n}\cdot \mathbf{b}$ is finite and $v\to +\infty$. Thus $q\to \pi /2$ with $p$ finite: the endpoint lies on ${\mathcal{I}}^{+}$. To the past, $v\to -\mathbf{n}\cdot \mathbf{b}$ remains finite while $u\to -\infty$, giving ${\mathcal{I}}^{-}$. Physical affine parameter is infinite at both ends. Along a radial null path, $dT=\pm dR$, so the diagram represents its null directions at $45$ degrees.

For a complete spacelike geodesic, take $t=t_0+a\lambda$, $\mathbf{x}=\mathbf{b}+\mathbf{w}\lambda$ with $|\mathbf{w}|>|a|$. At either end $r>|t|$ asymptotically, so $u\to -\infty$, $v\to +\infty$ and $(T,R)\to (0,\pi )$. Both ends therefore reach $i^0$, again at infinite physical affine parameter. These are the geodesic endpoints in Minkowski conformal compactification.

Only conformal preservation of null geodesic paths is guaranteed. The pictured timelike and spacelike curves are images of physical geodesics, not generally geodesics of the unphysical metric. For a null path, an unphysical affine parameter satisfies $d\tilde{\lambda }/d\lambda \propto {\Omega }^2$. Since $\Omega =O(1/|\lambda |)$ near a generic null-infinity endpoint, the unphysical parameter can converge while the physical parameter diverges. At $i^{\pm }$ and $i^0$, the usual smooth-boundary condition $d\Omega \ne 0$ degenerates; these vertices should not be treated as regular points of null infinity.

Complete Minkowski spacetime has no black-hole event horizon. Every physical event can send an outgoing null signal to ${\mathcal{I}}^{+}$, so the black-hole region $\mathcal{M}\setminus J^{-}({\mathcal{I}}^{+})$ is empty. There is likewise no particle or future event horizon for an eternal inertial observer: any finite event can eventually signal that observer. Null infinity is a conformal boundary, not a physical horizon. Uniformly accelerated observers can have Rindler horizons, but that is an observer-dependent restriction and is not a horizon of the full inertial Minkowski spacetime.

Differentiate the Friedmann equation and use the cosmological continuity equation to obtain $\dot{H}+H^2=-(4\pi G/3)(\rho +3P/c^2)$. The relation extends to turning points by continuity. Radiation obeys ${\rho }_R={\rho }_{R0}{a}^{-4}$, while dark energy with $P_{\Lambda }=-{\rho }_{\Lambda }{c}^2$ has constant density. Substitution at $a(t_0)=1$ gives $k{c}^2=\beta H_0^2$, and at arbitrary $a$ givesPut $x=a^2$. For $\beta >4$, the upward-opening polynomial $P(x)=x^2-\beta x+\beta$ has roots $x_{\pm }=(\beta \pm \sqrt{\beta (\beta -4)})/2$, with $1<x_{-}<2<x_{+}$. The expanding Big Bang branch reaches $x_{-}$ in finite time. There $H=0$ but $\dot{H}<0$, so expansion turns to contraction; it returns to $a=0$ in finite time, a Big Crunch. A sketch of $a^4{H}^2/H_0^2=P(x)$ has the forbidden negative interval between the roots, while $a^4\dot{H}/H_0^2=\beta (x-2)$ is a line crossing zero at two.

For $\beta =4$, $P(x)=(x-2{)}^2$. The branch starting at zero has $\dot{x}=2H_0(2-x)$, givingAt small $t$, $a\sim 2\sqrt{H_{0}t}$, the radiation-dominated scale factor. At large $t$, $a=\sqrt{2}[1-\frac{1}{2}{e}^{-2H_{0}t}+O(e^{-4H_{0}t})]$. It approaches an Einstein static universe with equal radiation and dark-energy densities, rather than recollapsing or entering unbounded expansion.

Use the original PDF's spatial Ricci component $R_{\chi \chi }=a\ddot{a}+2{\dot{a}}^2+2$; the converted TeX's $2\dot{a}\ddot{a}$ is a transcription defect. Continue with $c=1$ and signature $(-,+,+,+)$. Tracing the vacuum Einstein field equations with a cosmological constant gives $R=4\Lambda$, and hence $R_{ab}=\Lambda g_{ab}$. The time component therefore gives $\ddot{a}=(\Lambda /3)a$. Substituting this into the spatial component givesThe left side is strictly positive, so $\Lambda >0$ is necessary for a real positive closed scale factor. Define $\ell =\sqrt{3/\Lambda }$. The acceleration equation has solution $a=A{e}^{t/\ell }+B{e}^{-t/\ell }$, and the first integral fixes $4AB={\ell }^2$. Taking the positive scale-factor branch and shifting the origin to the minimum givesThe universe contracts for $t<0$, has its nonzero minimum $a=\ell$ at $t=0$, and expands for $t>0$. The minimum is a smooth bounce, not a curvature singularity.

Introduce the dimensionless time $\tau =t/\ell$. Then the closed slicing of de Sitter spacetime isOne can identify it explicitly: in five-dimensional Minkowski space, set $X^0=\ell \sinh \tau$ and $X^I=\ell \cosh \tau n^I$, where ${\sum }_{I=1}^4(n^I{)}^2=1$. These coordinates satisfy $-(X^0{)}^2+{\sum }_I(X^I{)}^2={\ell }^2$, and their induced metric is the one above. Thus the solution is global de Sitter spacetime, with spherical spatial topology and both contracting and expanding parts.

The time entering the printed compactification formula is this rescaled time. WriteDirect differentiation and elementary trigonometry give $dT/d\tau =\mathrm{sech}\tau$, $\sin T=\tanh \tau$ and $\cos T=\mathrm{sech}\tau$. ThereforeMultiplying by $(\cos T/\ell {)}^2$ gives the metric of a finite-time slab of the Einstein static universe, often called the Einstein cylinder. This is a conformal transformation of the metric and does not assert that the transformed matter content remains vacuum. The coordinate ranges areThe angular-coordinate degeneracies at the poles are the usual regular spherical ones.

Radial null lines satisfy $d\chi =\pm dT$, so the Penrose diagram is the square displayed below. Every interior point represents an angular two-sphere. Its top and bottom are the spacelike conformal infinities ${\mathcal{I}}^{+}$ and ${\mathcal{I}}^{-}$ at $T=\pi /2$ and $T=-\pi /2$; they are excluded from the physical spacetime and are not curvature singularities. The vertical sides $\chi =0$ and $\chi =\pi$ represent the regular north and south poles of the three-sphere, not spatial edges of the universe.

A particle horizon at an observer's current time bounds the comoving sources whose light could have reached that observer over the available past conformal interval. Here the cosmological past extends to $t=-\infty$, but its conformal duration is finite. For a comoving observer at $\chi =0$,Since $0<{\chi }_p<\pi$ at every finite time, a particle horizon exists. The corresponding null boundary in the diagram is $\chi =T+\pi /2$, emerging from the observer's past endpoint at ${\mathcal{I}}^{-}$. The antipodal point never has enough time to send a signal received at a finite proper time.

An observer event horizon bounds the events that can ever send a signal to the observer's future-inextendible worldline. The remaining conformal time fixes its distance:This too lies strictly between zero and $\pi$ at finite time, so the observer has a cosmological event horizon. Its null line ends at the observer's future endpoint on ${\mathcal{I}}^{+}$. Both horizon spheres have areal radius $a\sin \chi =\ell$, though their comoving angular positions differ except at $T=0$. These are the cosmological horizons in global de Sitter spacetime; changing the comoving observer relocates them by an isometry.

A Cauchy horizon is the boundary of the domain determined by data on a chosen partial Cauchy hypersurface; beyond it an inextendible causal curve can avoid that initial surface. Each complete constant-$T$ three-sphere here is a Cauchy hypersurface. To see this directly, a causal curve in the conformal metric has spatial speed at most one when parameterized by $T$. On any finite interior time interval its path therefore has a limit in the complete compact three-sphere. If an inextendible curve stopped at an interior time, that limit would allow its continuation, a contradiction. Its time range consequently extends between both conformal ends, and monotonicity makes it cross any complete constant-$T$ sphere exactly once. Its domain of dependence is the entire physical spacetime.

Thus particle and cosmological event horizons exist for every comoving observer, while the complete global Cauchy slices have no Cauchy horizons. Horizons of a restricted static patch or a deliberately partial initial surface are relative to that restriction; they do not alter the global Cauchy conclusion.

The radial part of the Minkowski metric becomes $-dudv$, and $r=(v-u)/2$. With $u=\tan p$, $v=\tan q$ and the positive conformal factor $\Omega =2\cos p\cos q$, multiplication by ${\Omega }^2$ givesConsequently the Minkowski conformal compactification isThe spatial metric $d{R}^2+{\sin }^{2}Rd{\Sigma }^2$ is the round unit three-sphere. The extended unphysical metric is therefore that of the Einstein static universe, $\mathbb{R}\times S^3$ with unit spatial radius. Physical Minkowski spacetime occupies only the diamond-shaped causal patch of this cylinder, or a triangle after restricting to nonnegative spherical radius; it is not the whole cylinder.

The complete coordinate ranges and joint restrictions areThe polar angles have the usual coordinate degeneracies at their poles and at $r=0$. In the extended Einstein static universe one may let $T\in \mathbb{R}$ and $0\le R\le \pi$, but these are not the ranges of the original physical patch. In its interior $\Omega >0$. The line $R=0$ is the regular centre $r=0$, not a component of infinity.

The conformal completion has future timelike infinity $i^{+}=(\pi ,0)$, past timelike infinity $i^{-}=(-\pi ,0)$, and spacelike infinity $i^0=(0,\pi )$. The open sloping boundaries arenamely future null infinity and past null infinity. Each ordinary interior point of the radial diagram represents a two-sphere, and each point on either open null boundary represents a two-sphere of null directions.

A timelike geodesic is an inertial worldline $\mathbf{x}=\mathbf{b}+\mathbf{\beta }t$ with $|\mathbf{\beta }|<1$. As $t\to +\infty$, both $t-r$ and $t+r$ tend to $+\infty$, giving $p,q\to \pi /2$ and endpoint $i^{+}$. At past infinity both tend to $-\infty$, giving $i^{-}$. These endpoints are reached at infinite physical proper time, although at finite conformal coordinate time $T$. A null geodesic has $r\sim |t|$. In its outgoing future, $u$ approaches a finite retarded time while $v\to +\infty$, giving an endpoint on ${\mathcal{I}}^{+}$; in its incoming past, $v$ has a finite limit and $u\to -\infty$, giving an endpoint on ${\mathcal{I}}^{-}$. A spacelike geodesic is a straight line with spatial tangent larger than its temporal tangent, so $u\to -\infty$, $v\to +\infty$ at either end and both endpoints are $i^0$ in the compactified radial description. All these physical geodesics are complete. Only the unparametrized null geodesics are necessarily preserved as geodesics of the unphysical metric; timelike and spacelike images need not be its geodesics.

For horizons, distinguish a property of the spacetime from a property of an observer or of an initial-data hypersurface. Complete inertial observers have no particle horizon: in the flat cosmological slicing $a(t)=1$, the past light-travel distance ${\int }_{-\infty }^{t_0}dt$ is infinite. There is no finite-age initial singularity truncating communication between comoving observers. This statement does not say that an event receives signals from spacelike-separated events at the same time.

There is also no black-hole event horizon, because every event can send a future-directed null ray to future null infinity. Nevertheless accelerated timelike observers can have both future and past observer event horizons. For an entire observer worldline $\gamma$, define them by $H_{\gamma }^{+}=\partial I^{-}(\gamma )$ and $H_{\gamma }^{-}=\partial I^{+}(\gamma )$, using its complete future and past respectively. An inertial observer's chronological past and future of the whole worldline are both all of Minkowski spacetime. By contrast, for the uniformly accelerated worldline in a Cartesian spatial direction,its null coordinates obey $t-x=-a^{-1}{e}^{-a\tau }<0$ and $t+x=a^{-1}{e}^{a\tau }>0$. An event with $t-x<0$ can signal to a sufficiently late point of this worldline, whereas an event with $t-x\ge 0$ cannot; finite transverse displacements do not change that conclusion as $t+x\to \infty$. Similarly the worldline can signal to precisely the events with $t+x>0$. Thus its two observer event horizons areThey are the Rindler horizons, and their accessible common region is $x>|t|$. They do not indicate a curvature singularity or a failure of global predictability.

Finally Minkowski spacetime is a globally hyperbolic spacetime: every inextendible causal curve crosses each complete spacelike plane $t=t_0$ exactly once. Indeed $t$ is strictly monotone along future-directed causal curves, and the speed bound $|d\mathbf{x}/dt|\le 1$ prevents an inextendible curve from ending at finite $t$ by escaping to spatial infinity. Its time range is therefore all of $\mathbb{R}$. These planes are Cauchy hypersurfaces with full domain of dependence, so there are no Cauchy horizons associated with complete Cauchy data.

A Cauchy horizon is, however, defined relative to a specified initial surface. Under a literal existence reading of “admits”, non-Cauchy initial surfaces can have such horizons even in flat spacetime. For example the edgeless spacelike hyperboloid ${\Sigma }_a:t=\sqrt{a^2+r^2}$ has $D({\Sigma }_a)=\{t>r\}$. To see this, inside the future light cone $t-r$ and $t+r$ are positive and nondecreasing on future-directed causal curves. Their product $t^2-r^2$ increases past $a^2$ on every future-inextendible curve from below the hyperboloid; every past-inextendible curve from above it crosses down through $a^2$ before leaving the cone. Outside the cone there is an inextendible null curve avoiding the hyperboloid. The future light cone itself is its past Cauchy horizon. The time-reversed hyperboloid gives a future Cauchy horizon. This is the dependence of a Cauchy horizon on the initial hypersurface, not an intrinsic loss of predictability of full Minkowski spacetime. Thus a categorical claim that no choice of partial initial surface can ever have a Cauchy horizon would be too strong.

For the Einstein static universe with $P=0$ and ${\rho }_0>0$, the acceleration equation gives $\Lambda =4\pi {\rho }_0$. The first Friedmann equation then gives $k/a_0^2=4\pi {\rho }_0$, soKeep $\Lambda ,k$ fixed and use the specified perturbation closure $\delta P=\delta \rho /3$. Linearization of the constraint gives $\delta \rho =-3{\rho }_0\delta a/a_0$. In the acceleration equation the background terms cancel, leavingThusThe growing mode establishes linear instability of the static universe; a specially chosen decaying perturbation does not remove instability. This uses the pressure-density relation explicitly supplied for the perturbation. If zero density is allowed, there is also the degenerate vacuum static case $k=0$, $\Lambda =0$, which is not the positive-density static universe considered here.

For $\Lambda >0$ and $k=1$,The unique maximum occurs atThe critical closed matter-cosmological-constant universe is obtained from $V_{\max }=-1$:

If $\Lambda >{\Lambda }_{\mathrm{crit}}$, then $V_{\max }<-1$: the energy line lies above the barrier, and a universe expanding from $a=0$ passes over it and expands forever toward de Sitter behavior. If $\Lambda ={\Lambda }_{\mathrm{crit}}$, the exact solution at $a=a_E$ is the Einstein static universe; the Big-Bang branch approaches it only asymptotically, while a separate branch can emerge from it and expand forever. If $0<\Lambda <{\Lambda }_{\mathrm{crit}}$, the equation $V(a)=-1$ has two positive roots $a_{-}<a_E<a_{+}$. The Big-Bang branch expands to $a_{-}$ and recollapses, while the disconnected large-$a$ branch contracts to $a_{+}$ and undergoes a Friedmann bounce for pressureless matter and positive cosmological constant. These are all qualitatively different possibilities.

In the orthonormal frame, the comoving perfect fluid in general relativity has $T_{00}=\rho$ and $T_{ij}=p{\delta }_{ij}$. The Einstein field equations with cosmological constant therefore giveThusChoosinggives $p=0$ and $\rho =1/(16\pi )>0$. The metric is then the Einstein static universe supported by pressureless matter and positive vacuum energy.